Question

Solve
$3a-16b+7c=03a+8b-5c=0a=?,b=?,c=?$.
By cross mutliplication.

Answer 1

Given $3a-16b+7c=0$,

$3a+8b-5c=0$

Ref. image

$\frac{a}{-16(-5)-7\left(8\right)}=\frac{b}{7\left(3\right)-(-5)3}=\frac{c}{3\left(8\right)-3(-16)}$

$\frac{a}{80-56}=\frac{b}{21+15}=\frac{c}{24+48}$

$\frac{a}{24}=\frac{b}{36}=\frac{c}{72}$

$(a,b,c)=(24,36,72)$

$\left(or\right)(a,b,c)=(2,3,6)$